Divisibility and Binomial Coefficients

[Max Alekseyev]

On Sun, May 4, 2008 at 5:12 AM, Stefan Steinerberger
<stefan.steinerberger at gmail.com> wrote:

> I find the "dual" question more interesting. Are there
> distinct a,b <= n such that C(n,a) + C(n,b) divides C(n,a+b)?

It seems that there is also restriction a+b <= n. Otherwise, there
would be a lot of solutions dividing C(n,a+b) = 0 when a+b>n.

> I only found the following seven solutions
>
> C(19,3)+C(19,5) | C(19,8)
> C(34,6)+C(34,7) | C(34,13)
> C(41,5)+C(41,7) | C(41,12)
> C(89,7)+C(89,8) | C(89,15)
> C(104,3)+C(104,4) | C(104,7)
> C(359,5)+C(359,6) | C(359,11)
> C(398,20)+C(398,21) | C(398,41)
>
> Is there any reason to assume that the number of solutions is
> finite/infinite?

Hard to say. But most of your solutions are of the form:

C(n,a)+C(n,a+1) | C(n,2a+1)

which is simply C(n+1,a+1) | C(n,2a+1).

The next solution of this form is

n = 2309, a = 5

Regards,
Max



Nice question. I have no good reason for saying so, but I would guess there 
are infinitely many solutions.

Here is a complete list for n <= 5000:


It might be worth trying to prove that a and b cannot differ by more than 2.

Drew

On May 4 2008, Stefan Steinerberger wrote:

>Dear seqfans,
>
>JM Bergot has asked me whether there are a < m,n
>such that C(m, a) + C(n, a) divides C(m+n,a), where
>C(m,n) is the binomial coefficient. As the nature of the
>problem suggests, there is a vast number of solutions
>(for example C(4,3) + C(5,3) = 14, C(9,3) = 6*14).
>
>I find the "dual" question more interesting. Are there
>distinct a,b <= n such that C(n,a) + C(n,b) divides C(n,a+b)?
>I only found the following seven solutions
>
>C(19,3)+C(19,5) | C(19,8)
>C(34,6)+C(34,7) | C(34,13)
>C(41,5)+C(41,7) | C(41,12)
>C(89,7)+C(89,8) | C(89,15)
>C(104,3)+C(104,4) | C(104,7)
>C(359,5)+C(359,6) | C(359,11)
>C(398,20)+C(398,21) | C(398,41)
>
>Is there any reason to assume that the number of solutions is
>finite/infinite?
>
>Best wishes,
>Stefan
>